Question: What is the extraneous solution to these equations? $\dfrac{x^2 + x}{x + 5} = \dfrac{-3x + 5}{x + 5}$
Answer: Multiply both sides by $x + 5$ $ \dfrac{x^2 + x}{x + 5} (x + 5) = \dfrac{-3x + 5}{x + 5} (x + 5)$ $ x^2 + x = -3x + 5$ Subtract $-3x + 5$ from both sides: $ x^2 + x - (-3x + 5) = -3x + 5 - (-3x + 5)$ $ x^2 + x + 3x - 5 = 0$ $ x^2 + 4x - 5 = 0$ Factor the expression: $ (x + 5)(x - 1) = 0$ Therefore $x = -5$ or $x = 1$ At $x = -5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -5$, it is an extraneous solution.